Question

Assertion: Pascal’s law is the working principle of a hydraulic lift.
Reason: Pressure is thrust per unit area.
A) Both assertion and reason are correct and reason is the correct explanation for assertion.
B) Both assertion and reason are correct but reason is not the correct explanation for assertion.
C) Assertion is correct but reason is incorrect
D) Assertion is incorrect but reason is correct.

Answer 1

Hint: In a hydraulic lift, a smaller pressure is applied to a volume of liquid with a small cross sectional area to a greater depth. This liquid transmits the pressure and helps to lift a load on the other end of the hydraulic lift with a larger cross sectional area but only to a shorter height.
Pascal’s law states that the pressure applied to an incompressible liquid is distributed equally, in all directions.

Formula used:
$\text{Pressure = }\dfrac{\text{Force}}{\text{Area}}$

$\dfrac{{{F}_{1}}}{{{A}_{1}}}=\dfrac{{{F}_{2}}}{{{A}_{2}}}$ --(Pascal’s Law)
Where, ${{F}_{1}}$ is the force applied to a liquid at a cross sectional area ${{A}_{1}}$. Since, this pressure is distributed equally according to Pascal’s law the ratio is equal to the ratio on the right hand side where ${{F}_{2}}$ is the force felt by another part of the liquid with cross sectional area ${{A}_{2}}$.

Complete step-by-step answer:
Pressure is the force or thrust (normal force) acting on a surface per unit area of the surface. Hence, mathematically, $\text{Pressure = }\dfrac{\text{Force}}{\text{Area}}$ ---(1)
Pascal’s law states that the pressure applied to an incompressible liquid is distributed equally, in all directions. That is, using (1),
$\text{Pressure applied at point 1 in a liquid = Pressure at point 2 in the liquid}$ Therefore,
$\dfrac{{{F}_{1}}}{{{A}_{1}}}=\dfrac{{{F}_{2}}}{{{A}_{2}}}$ --(Pascal’s Law) --(2)
Where, ${{F}_{1}}$ is the force applied to a liquid at a cross sectional area ${{A}_{1}}$. Since, this pressure is distributed equally according to Pascal’s law the ratio is equal to the ratio on the right hand side where ${{F}_{2}}$ is the force felt by another part of the liquid with cross sectional area ${{A}_{2}}$.
This principle is used in the principle of a hydraulic lift. A hydraulic lift has two ends connected together by a liquid medium. In a hydraulic lift a force ${{F}_{1}}$ is applied on an arm with the liquid having cross sectional area ${{A}_{1}}$. This pressure $\dfrac{{{F}_{1}}}{{{A}_{1}}}$ is transmitted through the liquid to the other arm with cross sectional area ${{A}_{2}}$ where $\left( {{A}_{2}}>{{A}_{1}} \right)$. Now, from (2), ${{F}_{2}}={{F}_{1}}\dfrac{{{A}_{2}}}{{{A}_{1}}}$ Now, since $\left( {{A}_{2}}>{{A}_{1}} \right)$, therefore $\left( \dfrac{{{A}_{2}}}{{{A}_{1}}}>1 \right)$ . Hence, $\left( {{F}_{2}}>{{F}_{1}} \right)$, that is, we can get a larger force on one arm by applying a lesser force on another arm.
This is the correct explanation of the working principle of a hydraulic lift.
Hence, both the assertion and reason are correct, but the reason is not the correct explanation for the assertion.
Hence, the correct option is B) Both assertion and reason are correct but reason is not the correct explanation for assertion.

Note: Students might be thinking that how can we get a larger force from a smaller force, so easily. This would mean that, in essence, we could have infinite force on one end and zero force on the other end and get free energy, thus solving the energy crisis of the world. However, that is not true.
One thing that has to be considered that since the liquid is incompressible, the volume remains constant. Hence, for a smaller cross sectional area we have to push the liquid to a greater depth and on the larger area side it rises to a smaller height. Since, the work done will be the product of the force and the height/depth, it will be the same on both sides since on the side where force is small, the depth is greater in the equal ratio that the force is large and the height is small on the other side. ${{F}_{1}}\times {{d}_{1}}={{F}_{2}}\times {{d}_{2}}$ $\text{Work done = }F\times d$ Therefore, the work done remains equal on both sides and we are not getting free energy.